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Graduate Aptitude Test in Engineering

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Engineering Mathematics

General Aptitude

1

A plane bisects the line segment joining the points (1, 2, 3) and ($$-$$ 3, 4, 5) at rigt angles. Then this plane also passes through the point :

A

($$-$$ 3, 2, 1)

B

(3, 2, 1)

C

($$-$$ 1, 2, 3)

D

(1, 2, $$-$$ 3)

Since the plane bisects the line joining the points (1, 2, 3) and ($$-$$3, 4, 5) then the plane passes through the midpoint of the line which is :

$$\left( {{{1 - 3} \over 2},{{2 + 4} \over 2},{{5 + 3} \over 2}} \right)$$ $$ \equiv $$ $$\left( {{{ - 2} \over 2},{6 \over 2},{8 \over 2}} \right)$$ $$ \equiv $$ ($$-$$1, 3, 4).

As plane cuts the line segment at right angle, so the direction cosines of the normal of the plane are ($$-$$ 3 $$-$$ 1, 4 $$-$$ 2, 5 $$-$$ 3) = ($$-$$ 4, 2, 2)

So the equation of the plane is : $$-$$ 4x + 2y + 2z = $$\lambda $$

As plane passes through ($$-$$ 1, 3, 4)

So, $$-$$ 4($$-$$ 1) + 2(3) + 2(4) = $$\lambda $$ $$ \Rightarrow $$ $$\lambda $$ = 18

Therefore, equation of plane is : $$-$$ 4x + 2y + 2z = 18

Now, only ($$-$$ 3, 2, 1) satiesfies the given plane as

$$-$$ 4($$-$$ 3) + 2(2) + 2(1) = 18

$$\left( {{{1 - 3} \over 2},{{2 + 4} \over 2},{{5 + 3} \over 2}} \right)$$ $$ \equiv $$ $$\left( {{{ - 2} \over 2},{6 \over 2},{8 \over 2}} \right)$$ $$ \equiv $$ ($$-$$1, 3, 4).

As plane cuts the line segment at right angle, so the direction cosines of the normal of the plane are ($$-$$ 3 $$-$$ 1, 4 $$-$$ 2, 5 $$-$$ 3) = ($$-$$ 4, 2, 2)

So the equation of the plane is : $$-$$ 4x + 2y + 2z = $$\lambda $$

As plane passes through ($$-$$ 1, 3, 4)

So, $$-$$ 4($$-$$ 1) + 2(3) + 2(4) = $$\lambda $$ $$ \Rightarrow $$ $$\lambda $$ = 18

Therefore, equation of plane is : $$-$$ 4x + 2y + 2z = 18

Now, only ($$-$$ 3, 2, 1) satiesfies the given plane as

$$-$$ 4($$-$$ 3) + 2(2) + 2(1) = 18

2

If the position vectors of the vertices A, B and C of a $$\Delta $$ ABC are respectively $$4\widehat i + 7\widehat j + 8\widehat k,$$ $$2\widehat i + 3\widehat j + 4\widehat k,$$ and $$2\widehat i + 5\widehat j + 7\widehat k,$$ then the position vectors of the point, where the bisector of $$\angle $$A meets BC is :

A

$${1 \over 2}\left( {4\widehat i + 8\widehat j + 11\widehat k} \right)$$

B

$${1 \over 3}\left( {6\widehat i + 11\widehat j + 15\widehat k} \right)$$

C

$${1 \over 3}\left( {6\widehat i + 13\widehat j + 18\widehat k} \right)$$

D

$${1 \over 4}\left( {8\widehat i + 14\widehat j + 19\widehat k} \right)$$

Suppose angular bisector of A meets BC at D(x, , z)

Using angular bisector theorem,

$${{AB} \over {AC}}$$ = $${{BD} \over {DC}}$$

$${{BD} \over {DC}}$$ = $${{\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 3} \right)}^2} + {{\left( {8 - 4} \right)}^2}} } \over {\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 5} \right)}^2} + {{\left( {8 - 7} \right)}^2}} }}$$

= $${{\sqrt {{2^2} + {4^2} + {4^2}} } \over {\sqrt {{2^2} + {2^2} + {1^2}} }}$$ = $${6 \over 3}$$ = 2

So, D(x, y, z) $$ \equiv $$ $$\left( {{{\left( 2 \right)\left( 2 \right) + \left( 1 \right)\left( 2 \right)} \over {2 + 1}},{{\left( 2 \right)\left( 5 \right) + \left( 1 \right)\left( 3 \right)} \over {2 + 1}}} \right.$$

$$\left. {{{\left( 2 \right)\left( 7 \right) + \left( 1 \right)\left( 4 \right)} \over {2 + 1}}} \right)$$

D(x, y, z) $$ \equiv $$ $$\left( {{6 \over 3},{{13} \over 3},{{18} \over 3}} \right)$$

Therefore, position vector of point p = $${1 \over 3}$$ (6i + 13j + 18k)

Using angular bisector theorem,

$${{AB} \over {AC}}$$ = $${{BD} \over {DC}}$$

$${{BD} \over {DC}}$$ = $${{\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 3} \right)}^2} + {{\left( {8 - 4} \right)}^2}} } \over {\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 5} \right)}^2} + {{\left( {8 - 7} \right)}^2}} }}$$

= $${{\sqrt {{2^2} + {4^2} + {4^2}} } \over {\sqrt {{2^2} + {2^2} + {1^2}} }}$$ = $${6 \over 3}$$ = 2

So, D(x, y, z) $$ \equiv $$ $$\left( {{{\left( 2 \right)\left( 2 \right) + \left( 1 \right)\left( 2 \right)} \over {2 + 1}},{{\left( 2 \right)\left( 5 \right) + \left( 1 \right)\left( 3 \right)} \over {2 + 1}}} \right.$$

$$\left. {{{\left( 2 \right)\left( 7 \right) + \left( 1 \right)\left( 4 \right)} \over {2 + 1}}} \right)$$

D(x, y, z) $$ \equiv $$ $$\left( {{6 \over 3},{{13} \over 3},{{18} \over 3}} \right)$$

Therefore, position vector of point p = $${1 \over 3}$$ (6i + 13j + 18k)

3

The sum of the intercepts on the coordinate axes of the plane passing through the point ($$-$$2, $$-2,$$ 2) and containing the line joining the points (1, $$-$$1, 2) and (1, 1, 1) is :

A

4

B

$$-$$ 4

C

$$-$$ 8

D

12

Equation of plane passing through three given points is :

$$\left| {\matrix{ {x - {x_1}} & {y - {y_1}} & {z - {z_1}} \cr {{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr {{x_3} - {x_1}} & {{y_3} - {y_1}} & {{z_3} - {z_1}} \cr } } \right| = 0$$

$$ \Rightarrow $$ $$\left| {\matrix{ {x + 2} & {y + 2} & {z - 2} \cr {1 + 2} & { - 1 + 2} & {2 - 2} \cr {1 + 2} & {1 + 2} & {1 - 2} \cr } } \right| = 0$$

$$ \Rightarrow $$ $$\left| {\matrix{ {x + 2} & {y + 2} & {z - 2} \cr 3 & 1 & 0 \cr 3 & {30} & { - 1} \cr } } \right| = 0$$

$$ \Rightarrow $$ $$ - x + 3y + 6z - 8 = 0$$

$$ \Rightarrow $$ $${x \over 8} - {{3y} \over 8} - {{6z} \over 8} + {8 \over 8} = 0$$

$$ \Rightarrow $$ $${x \over 8} - {y \over {{8 \over 3}}} - {z \over {{8 \over 6}}} = - 1$$

$$ \Rightarrow $$ $${x \over { - 8}} + {y \over {{8 \over 3}}} + {z \over {{8 \over 6}}} = 1$$

$$ \therefore $$ Sum of intercepts $$ = - 8 + {8 \over 3} + {8 \over 6} = - 4$$

$$\left| {\matrix{ {x - {x_1}} & {y - {y_1}} & {z - {z_1}} \cr {{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr {{x_3} - {x_1}} & {{y_3} - {y_1}} & {{z_3} - {z_1}} \cr } } \right| = 0$$

$$ \Rightarrow $$ $$\left| {\matrix{ {x + 2} & {y + 2} & {z - 2} \cr {1 + 2} & { - 1 + 2} & {2 - 2} \cr {1 + 2} & {1 + 2} & {1 - 2} \cr } } \right| = 0$$

$$ \Rightarrow $$ $$\left| {\matrix{ {x + 2} & {y + 2} & {z - 2} \cr 3 & 1 & 0 \cr 3 & {30} & { - 1} \cr } } \right| = 0$$

$$ \Rightarrow $$ $$ - x + 3y + 6z - 8 = 0$$

$$ \Rightarrow $$ $${x \over 8} - {{3y} \over 8} - {{6z} \over 8} + {8 \over 8} = 0$$

$$ \Rightarrow $$ $${x \over 8} - {y \over {{8 \over 3}}} - {z \over {{8 \over 6}}} = - 1$$

$$ \Rightarrow $$ $${x \over { - 8}} + {y \over {{8 \over 3}}} + {z \over {{8 \over 6}}} = 1$$

$$ \therefore $$ Sum of intercepts $$ = - 8 + {8 \over 3} + {8 \over 6} = - 4$$

4

If the angle between the lines, $${x \over 2} = {y \over 2} = {z \over 1}$$

and $${{5 - x} \over { - 2}} = {{7y - 14} \over p} = {{z - 3} \over 4}\,\,$$ is $${\cos ^{ - 1}}\left( {{2 \over 3}} \right),$$ then p is equal to :

and $${{5 - x} \over { - 2}} = {{7y - 14} \over p} = {{z - 3} \over 4}\,\,$$ is $${\cos ^{ - 1}}\left( {{2 \over 3}} \right),$$ then p is equal to :

A

$${7 \over 2}$$

B

$${2 \over 7}$$

C

$$-$$ $${7 \over 4}$$

D

$$-$$ $${4 \over 7}$$

Let $$\theta $$ be the angle between the two lines

Here direction cosines of $${x \over 2}$$ = $${y \over 2}$$ = $${z \over 1}$$ are 2, 2, 1

Also second line can be written as :

$${{x - 5} \over 2} = {{y - 2} \over {{P \over 7}}} = {{z - 3} \over 4}$$

$$ \therefore $$ its direction cosines are 2, $${{P \over 7}}$$, 4

Also, cos$$\theta $$ = $${2 \over 3}$$ (Given)

$$ \because $$ cos$$\theta $$ $$ = \left| {{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}} \over {\sqrt {a_1^2 + b_1^2 + c_1^2\sqrt {a_2^2 + b_2^2 + c_2^2} } }}} \right|$$

$$ \Rightarrow $$ $${2 \over 3}$$ $$ = \left| {{{\left( {2 \times 2} \right) + \left( {2 \times {P \over 7}} \right) + \left( {1 \times 4} \right)} \over {\sqrt {{2^2} + {2^2} + {1^2}} \sqrt {{2^2} + {{{P^2}} \over {49}} + {4^2}} }}} \right|$$

$$ = {{4 + {{2P} \over 7} + 4} \over {3 \times \sqrt {{2^2} + {{{P^2}} \over {49}} + {4^2}} }}$$

$$ \Rightarrow $$ $${\left( {4 + {P \over 7}} \right)^2} = 20 + {{{P^2}} \over {49}}$$

$$ \Rightarrow $$ 16 + $${{8P} \over 7} + {{{P^2}} \over {49}}$$ = 20 + $${{{P^2}} \over {49}}$$

$$ \Rightarrow $$ $${{8P} \over 7} = 4$$ $$ \Rightarrow $$ $$P = {7 \over 2}$$

Here direction cosines of $${x \over 2}$$ = $${y \over 2}$$ = $${z \over 1}$$ are 2, 2, 1

Also second line can be written as :

$${{x - 5} \over 2} = {{y - 2} \over {{P \over 7}}} = {{z - 3} \over 4}$$

$$ \therefore $$ its direction cosines are 2, $${{P \over 7}}$$, 4

Also, cos$$\theta $$ = $${2 \over 3}$$ (Given)

$$ \because $$ cos$$\theta $$ $$ = \left| {{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}} \over {\sqrt {a_1^2 + b_1^2 + c_1^2\sqrt {a_2^2 + b_2^2 + c_2^2} } }}} \right|$$

$$ \Rightarrow $$ $${2 \over 3}$$ $$ = \left| {{{\left( {2 \times 2} \right) + \left( {2 \times {P \over 7}} \right) + \left( {1 \times 4} \right)} \over {\sqrt {{2^2} + {2^2} + {1^2}} \sqrt {{2^2} + {{{P^2}} \over {49}} + {4^2}} }}} \right|$$

$$ = {{4 + {{2P} \over 7} + 4} \over {3 \times \sqrt {{2^2} + {{{P^2}} \over {49}} + {4^2}} }}$$

$$ \Rightarrow $$ $${\left( {4 + {P \over 7}} \right)^2} = 20 + {{{P^2}} \over {49}}$$

$$ \Rightarrow $$ 16 + $${{8P} \over 7} + {{{P^2}} \over {49}}$$ = 20 + $${{{P^2}} \over {49}}$$

$$ \Rightarrow $$ $${{8P} \over 7} = 4$$ $$ \Rightarrow $$ $$P = {7 \over 2}$$

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